Here, you will practice factoring trinomials of the form
$\,x^2 + bx + c\,$,
where $\,b\,$ and $\,c\,$ are integers.
Notice that the coefficient of the $\,x^2\,$ term is $\,1\,$.
Recall that the integers are:
$\,\ldots,3,2,1,0,1,2,3,\ldots$
As discussed in Basic Concepts Involved in Factoring Trinomials,
you must first find two numbers that add to $\,b\,$ and that multiply to $\,c\,$, since then:
$$
\,x^2 + bx + c \ \ =\ \ x^2 + (\overset{=\ b}{\overbrace{f+g}})x + \overset{=\ c}{\overbrace{\ fg\ }} \ \ =\ \ (x + f)(x + g)
$$
As discussed in
Factoring Trinomials of the form
$\,x^2 + bx + c\,$, where $c\gt 0$:
if $\,c\,$ is positive, then both numbers will be positive, or both numbers will be negative.
When you add numbers that have the same sign, then in your head you do an addition problem.
As discussed in
Factoring Trinomials of the form
$\,x^2 + bx + c\,$, where $c\lt 0$:
if $\,c\,$ is negative, then one number will be positive, and the other will be negative.
When you add numbers that have different signs, then in your head you do a subtraction problem.
the ‘PANS’ memory device
When you're trying to find the two numbers that work,
you always want to do the mental arithmetic with only positive numbers.
It's much easier this way.
Here are the steps:
 Check that the coefficient of the square term is $\,1\,$.
 Look at the constant term ($\,c\,$).
This is the number that tells you whether, in your head, you'll think of adding or subtracting.
Is it Positive? Then, you'll be Adding.
Is it Negative? Then, you'll be Subtracting.
This is the PANS memory device!
Positive Addition Negative Subtraction
 Throw away all the minus signs (if any)
and find two numbers that multiply to
$\,c\,$
and add/subtract (as appropriate) to $\,b\,$.
Lots of examples below.
 Adjust the sign(s) of your two numbers, as needed.
(Read the details about this in the earlier web exercises.)
 Use your two numbers to factor the trinomial, as illustrated in the examples below.
 Be sure to check your answer using FOIL.
EXAMPLES:
Factor:
$x^2 + 5x + 6$
Solution:
 Is the coefficient of the
$\,x^2\,$ term equal to $\,1\,$? Check!
 Look at the constant term, $\,6\,$.
It's positive.
PANS—Positive, Add.
 There aren't any minus signs to throw away.
Find two numbers that multiply to $\,6\,$ and add to $\,5\,$.
Got them? $\,2\,$ and $\,3\,$.
 That $\,5\,$ in the middle is positive.
So, both numbers will be positive.
The two numbers are $\,2\,$ and $\,3\,$.

$\displaystyle
\,x^2 + 5x + 6 \ \ =\ \ x^2 + (\overset{=\ 5}{\overbrace{2 + 3}})x + \overset{=\ 6}{\overbrace{\ 2\cdot 3\ }} \ \ =\ \ (x + 2)(x + 3)
$
Factor:
$x^2 + 5x  6$
Solution:
 Is the coefficient of the
$\,x^2\,$ term equal to $\,1\,$? Check!
 Look at the constant term, $\,6\,$.
It's negative.
PANS—Negative, Subtract.
 Throw away all minus signs.
Find two numbers that multiply to $\,6\,$ and subtract to give $\,5\,$.
Got them? $\,6\,$ and $\,1\,$.
 That $\,5\,$ in the middle is positive.
So, the bigger number will be positive.
The two numbers are $\,6\,$ and $\,1\,$.

$\displaystyle
\,x^2 + 5x  6 \ \ =\ \ x^2 + (\overset{=\ 5}{\overbrace{6 + (1)}})x + \overset{=\ 6}{\overbrace{\ 6\cdot(1)\ }} \ \ =\ \ (x + 6)(x  1)
$
Factor:
$x^2  5x + 6$
Solution:
 Is the coefficient of the
$\,x^2\,$ term equal to $\,1\,$? Check!
 Look at the constant term, $\,6\,$.
It's positive.
PANS—Positive, Add.
 Throw away all minus signs.
Find two numbers that multiply to $\,6\,$ and add to $\,5\,$.
Got them? $\,2\,$ and $\,3\,$.
 That $\,5\,$ in the middle is negative.
So, both numbers will be negative.
The two numbers are $\,2\,$ and $\,3\,$.

$\displaystyle
\,x^2  5x + 6 \ \ =\ \ x^2 + (\overset{=\ 5}{\overbrace{(2) + (3)}})x + \overset{=\ 6}{\overbrace{\ (2)\cdot (3)\ }} \ \ =\ \ (x  2)(x  3)
$
Factor:
$x^2  5x  6$
Solution:
 Is the coefficient of the
$\,x^2\,$ term equal to $\,1\,$? Check!
 Look at the constant term, $\,6\,$.
It's negative.
PANS—Negative, Subtract.
 Throw away all minus signs.
Find two numbers that multiply to $\,6\,$ and subtract to give $\,5\,$.
Got them? $\,6\,$ and $\,1\,$.
 That $\,5\,$ in the middle is negative.
So, the bigger number will be negative.
The two numbers are $\,6\,$ and $\,1\,$.

$\displaystyle
\,x^2  5x  6 \ \ =\ \ x^2 + (\overset{=\ 5}{\overbrace{(6) + 1}})x + \overset{=\ 6}{\overbrace{\ (6)\cdot 1\ }} \ \ =\ \ (x  6)(x + 1)
$
Factor:
$x^2 + 3x  7$
Solution:
 Is the coefficient of the $\,x^2\,$ term equal to $\,1\,$? Check!
 Look at the constant term, $\,7\,$.
It's negative.
PANS—Negative, Subtract.
 Throw away all minus signs.
Find two numbers that multiply to $\,7\,$ and subtract to give $\,3\,$.
There aren't any.
$x^2 + 3x  7\,$ is not factorable over the integers.
Master the ideas from this section
by practicing the exercise at the bottom of this page.
When you're done practicing, move on to:
Identifying Quadratic Equations
For more advanced students, a graph is displayed.
For example, suppose you're asked to factor
$\,x^2 + 5x  6\,$.
Then, you'll see the graph of
$\,y = x^2 + 5x  6\,$.
Pay attention to where it crosses the $\,x\,$axis (at $\,6\,$ and $\,1\,$).
Compare it with the factorization:
$x^2 + 5x  6 = (x + 6)(x  1)\,$
See any relationship?
You're discovering the beautiful relationship between the zeroes of a polynomial, and its factors!
Click the “show/hide graph” button if you prefer not to see the graph.
If a trinomial is not factorable over the integers, input ‘NF’ (for ‘Not Factorable’).